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No: Take $k$ to be the rational numbers and $G$ to be the group of third roots of unity. Then the only rational point in $G$ is $1$. Then take $H$ to be the component of the identity. This satisfies your conditions but $H \neq G$.

The problem here is that the groups are no connected.

edit: I notice now after posting my answer that David Speyer suggested the same example in a similar question...

Ana, as I explained in my comment to your previous question, in characteristic zero the rational points are always Zariski-dense in a connected linear algebraic group. (This is not obvious, and requires some serious input from the structure theory of such groups, both the reductive case and the general case.) Hence, the answer to your question is affirmative.
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BCnrdJul 15 '10 at 17:29